Lecture 6 Examples and Problems

Lecture 6 Examples and Problems Heat capacity of solids & liquids Thermal diffusion Thermal conductivity Irreversibility Hot Cold Random Walk and Particle Diffusion Counting and Probability Microstates and Macrostates The meaning of equilibrium Ω(m) 9 spins m -9-7 -5-3 -1 1 3 5 7 9 Lecture 6, p. 1

Heat Capacity and Specific Heat Heat capacity: The heat energy required to raise the temperature of an object by 1K (=1º C). It depends on the amount of material. Units: J / K C Q = T Upper case C Specific heat: The heat capacity normalized to a standard amount of material (mass or moles). It only depends on the kind of material. Normalize to mass: Units: J/kg. K c = C m Lower case c Normalize to moles: molar specific heat Units: J/mole. K c mol = C n Lecture 6, p.

Act 1 1 calorie* is defined to be the energy needed to raise the temperature of 1 gram of water 1 C (= 1 K). Therefore, given that c HO = 4184 J/kg. K, 1 calorie = 4.18 Joule. Upper case C lower case c 1 food Calorie = 1000 calorie = 4180 J If you weigh 80 kg, consume 000 Cal/day, and could actually convert this entirely into work, how high could you climb? a. 1 km b. 10 km c. 100 km *The amount of heat actually depends somewhat on the temperature of the water, so there are actually several slightly different calorie definitions. Lecture 6, p. 3

Solution 1 calorie is defined to be the energy needed to raise the temperature of 1 gram of water 1 C (= 1 K). Therefore, given that c HO = 4184 J/kg. K, 1 calorie = 4.18 Joule. Upper case C lower case c 1 food Calorie = 1000 calorie = 4180 J If you weigh 80 kg, consume 000 Cal/day, and could actually convert this entirely into work, how high could you climb? a. 1 km b. 10 km c. 100 km 000 Cal = 8.36 10 6 J = mgh So, h = 8.36 10 6 J / (80 kg 9.8 m/s ) = 10.7 km Note: Tour de France riders consume 6000-9000 Cal/day. Lecture 6, p. 4

Exercise: Spacecraft Heat Shields This false-color view of Titan (moon of Saturn) is a composite of images captured by Cassini's infrared camera, which can penetrate some of Titan's clouds. Light and dark regions in the upper left quadrant are unknown types of terrain on Titan's surface. The Huygens spacecraft entered the atmosphere on Jan. 14, 005, initially traveling at ~6 km/s. After decelerating from friction, the heat shield was jettisoned, and three parachutes were deployed to allow a soft landing. What is the temperature rise on entry, assuming that half of the thermal energy goes into the ship (and half to the atmosphere)? Assume c steel = 500 J/kg-K. Lecture 6, p. 5

Solution v = 6 km/s c = 500 J/kg-K Use conservation of energy (1 st law of thermodynamics). Half of the initial kinetic energy becomes internal thermal energy. 1 1 mv = C T = cm T 3 v (6 10 m/s) T = = = 18,000 K! 4c 4 500 J/kg-K Note that m cancels. The problem is that steel melts at ~1700 K! For this reason, the heat shield is not made of steel, but rather a ceramic that burns off ( ablates ). Also, the ceramic has a very low thermal conductivity! Lecture 6, p. 6

Exercise: Heat Capacity Two blocks of the same material are put in contact. Block 1 has m 1 = 1 kg, and its initial temperature is T 1 = 75 C. Block has m = kg, and T = 5 C. What is the temperature after the blocks reach thermal equilibrium? 1 Lecture 6, p. 7

Solution Two blocks of the same material are put in contact. Block 1 has m 1 = 1 kg, and its initial temperature is T 1 = 75 C. Block has m = kg, and T = 5 C. What is the temperature after the blocks reach thermal equilibrium? 1 The two blocks have the same (unknown) specific heat. However, the heat capacity of block, C = cm, is twice as large as that of block 1, C 1 = cm 1. We can use the 1st law (conservation of energy) to determine the final temperature: U f = c(m 1 +m )T f = U i = cm 1 T 1 + cm T. Solve for T f : T f mt m + m T 1 75 + 5 + m 3 1 1 = = = 1 1 41.7 C Question: Suppose we measure temperature in Kelvin. Will we get a different answer? Lecture 6, p. 8

Heat conduction How long does it take? For simplicity we assume that system B is really big (a thermal reservoir ), so that it s temperature is always T B. T A (t) T B T A0 T A (t) = T B + T(t) T B C U = T Heat into A: TA ( t) TB T ( t) QA du A d( T ( t)) H A (t) = = = = = CA R th R th dt dt dt d( T ( t)) T ( t) = dt R C th A Guess solution: T(t) = T(t=0) e -t/τ, or T A (t) = T B + (T A0 T B ) e -t/τ Plug into above DiffEQ: τ = R th C A (like a discharging capacitor!) Lecture 6, p. 9

Act : Exponential Cooling A hot steel bearing (at T = 00 C) is dropped into a large vat of cold water at 10 C. Compare the time it takes the bearing to cool from 00 to 190 C to the time it takes to cool from 100 to 90 C.(Assume the specific heat of steel is ~constant over this temperature range.) a. t 00 190C > t 100 90C b. t 00 190C = t 100 90C c. t 00 190C < t 100 90C Lecture 6, p. 10

Solution A hot steel bearing (at T = 00 C) is dropped into a large vat of cold water at 10 C. Compare the time it takes the bearing to cool from 00 to 190 C to the time it takes to cool from 100 to 90 C.(Assume the specific heat of steel is ~constant over this temperature range.) a. t 00 190C > t 100 90C b. t 00 190C = t 100 90C c. t 00 190C < t 100 90C However, the rate of heat flow out of the bearing depends on T bearing (t) T water, and is different (~190 C and ~90 C) for the two cases. Because more heat flows at the outset, the initial temperature drop is faster. 00C t 00 190C t 100 90C 10C Lecture 6, p. 11

FYI: Thermal Diffusion and Heat Conduction How is it that random motion can give heat flow in a particular direction? Thermal energy randomly diffuses around, spreading out. However, the heat flow out of a region is proportional to the amount of energy that is there at that time. Look at region. More heat will randomly diffuse in from a high T region than from low T: J 1 > J 1, and J 3 > J 3. So there will be a net flow of heat in the direction of decreasing T. In 1-D the heat current density is: J H dt = κ dx where κ is the thermal conductivity, a property of the material. T 1 3 J 1 J 1 J 3 J 3 x Lecture 6, p. 1

The Diffusion Constant The mean square displacement along x is still proportional to t: l 1 x = Dt, where D = = vl These are the average 3 values of τ, v, and l. 3τ D is called the diffusion constant. xrms = < x > = Dt The 3-D displacement (along x, y, and z combined) is: r = x + y + z = 6Dt Lecture 5, p 13

Exercise: Diffusion A perfume bottle is opened, and the molecules begin to diffuse through the air. Suppose that perfume molecules move about 10 nm between collisions with air molecules, and that their average speed is v ~ 100 m/s. 1) What is the diffusion constant of perfume in air? ) Suppose you hold your nose 10 cm from the perfume bottle. When will you be able to smell the perfume? What about a person across the room (5 m away)? 3) Are these times reasonable? Lecture 6, p. 14

Solution A perfume bottle is opened, and the molecules begin to diffuse through the air. Suppose that perfume molecules move about 10 nm between collisions with air molecules, and that their average speed is v ~ 100 m/s. 1) What is the diffusion constant of perfume in air? D = vl = 1 7 3.33 10 m /s 3 ) Suppose you hold your nose 10 cm from the perfume bottle. When will you be able to smell the perfume? What about a person across the room (5 m away)? t = x D = = 4 1.5 10 s ~ 4 hours (nose) 7 3.8 10 s ~ 1 year (across room) 3) Are these times reasonable? These times are much too long. In this situation, conduction (air currents) is much more important than diffusion. However, in solids, where conduction is rarely important, diffusion can dominate. Lecture 6, p. 15

Act 3: Isotope Separation You have the task of separating two isotopes of Uranium: 35 U and 38 U. Your lab partner suggests the following: Put a gas containing both of them at one end of a long tube through which they will diffuse. Which will get to the far end first? A) 35 U B) 38 U C) Neither (equal time) Lecture 6, p. 16

Solution You have the task of separating two isotopes of Uranium: 35 U and 38 U. Your lab partner suggests the following: Put a gas containing both of them at one end of a long tube through which they will diffuse. Which will get to the far end first? A) 35 U B) 38 U C) Neither (equal time) The diffusion time t ~ l /3D, where D = vl/3. From equipartition: 1 3 mv = kt v = 3 kt / m Therefore t 1/D 1/v m. The heavier isotope takes slightly longer. (This is the technique first used in the Manhattan Project. It was then found that centrifuges speed up the process.) Lecture 6, p. 17

The Meaning of Equilibrium (1) Definitions: Macrostate: Microstate: Assume: An Introduction to Statistical Mechanics The set of quantities we are interested in (e.g., p, V, T). A specific internal configuration of the system, with definite values of all the internal variables. Due to the randomness of thermal processes, every microstate is equally likely. Therefore the probability of observing a particular macrostate is proportional to the number of corresponding microstates. P(A) = Ω Α /Ω tot Many systems are described by binary distributions: Random walk Coin flipping This can be solved using the binomial formula, Electron spin because each particle has two choices: N! Ω ( N,N L ) = NCN L N! N N! Ω = tot N L ( ) L Lecture 6, p. 18

Example: Electron Spin Electrons have spin and associated magnetic moment µ. They can only point up or down :* or One microstate Consider a system of N=9 spins: Similar to HHTTHTHHT in a coin flipping experiment: The total magnetic moment (what we can measure) is: M = (N up - N down )µ mµ m = spin excess = N up N down A macrostate is described by m. The microstate above has m = +1. *This is a result from P14 that you ll have to take on faith. Lecture 6, p. 19

Electron Spin () Count microstates for each value of m: N! N! Ω (N up) = Ω (m) = N N m N m up!n down! +!! = N! 1 P(N up) N N!N! up down Ω(N up ) or Ω(m) Number of up spins: N up = 0 1 3 4 5 6 7 8 9 N up - N down : m = -9-7 -5-3 -1 1 3 5 7 9 # microstates: Ω = 1 9 36 84 16 16 84 36 9 1 Each macrostate is described by m. (what we measure). This problem will become more interesting later in the course. We will put the spins in a magnetic field. The energy of spin up will not equal the spin down energy, and the probabilities will change. Lecture 6, p. 0

Gaussian Approximation to the Binomial Distribution When N is large, the binomial formula becomes impossible to evaluate on a calculator. N! N! 1 Ω (N ) = P(N ) = up up N N up!n down! N up!n down! Fortunately, when N is large the shape of the distribution becomes a Gaussian: m / N ( ) ( ) 1/ 1/ (m) N e P(m) e Ω = = πn m / N This expression can be evaluated for very large N (e.g., 10 3 ). πn Gaussian: Does it work? Try N = 0, m = : Binomial: Ω(1) = 167960, P(1) = 0.160 Gaussian: Ω(1) = 16976, P(1) = 0.161 The agreement improves as N increases. Ex: N = 10 10, m = 10 5 P(m) = 8.0 10-6 Lecture 6, p. 1

Exercise: Random Walk Suppose that a particle is undergoing a 1-dimensional random walk (equally likely steps in the + or minus directions.) What is the probability: 1) that after N steps it is exactly where it started? Evaluate it for N=10. ) that after N steps it is within steps of the maximum possible positive position? Evaluate it for N=10. Lecture 6, p.

Solution Suppose that a particle is undergoing a 1-dimensional random walk (equally likely steps in the + or minus directions.) What is the probability: 1) that after N steps it is exactly where it started? Evaluate it for N=10. There are N total microstates. We want N + = N - = N/. The probability that this happens is: ( ) N! P 0 = = 0.46 N! N ( ) ) that after N steps it is within steps of the maximum possible positive position? Evaluate it for N=10. Lecture 6, p. 3

Solution Suppose that a particle is undergoing a 1-dimensional random walk (equally likely steps in the + or minus directions.) What is the probability: 1) that after N steps it is exactly where it started? Evaluate it for N=10. There are N total microstates. We want N + = N - = N/. The probability that this happens is: ( ) N! P 0 = = 0.46 N! N ( ) ) that after N steps it is within steps of the maximum possible positive position? Evaluate it for N=10. We must sum the three probabilities: P(N) + P(N-1) + P(N-). P(N) = 1/ N one microstate P(N-1) = 0 N + - N - must be even when N is even. P(N-) = N/ N N microstates: N + = N-1 and N - = 1. The sum is ( ) N + 1 P N : N = = 0.011 N Lecture 6, p. 4

Exercise: Random Walk () Do the previous exercise when N = 10 6. For what value of m is P(m) half of P(0)? Lecture 6, p. 5

Solution Do the previous exercise when N = 10 6. There is no way that we are going to evaluate 1,000,000. We must use the Gaussian approximation: P(m) = ( ) 1/ m / N e πn 1) m = 0: ( ) ( 6 ) P 0 π10 1/ = = 8 10 4 ) m = N and m = N-: 1/ 6 10 / P N = e = 0 for all practical purposes. π10 ( ) ( 6 ) For what value of m is P(m) half of P(0)? We want e = m = ln( ) N = 1.177 N = 1177 m / N 1 Lecture 6, p. 6

Example: Probability & Microstates The typical baseball player gets a hit 5% of the time. If this player gets several hits in a row, he is said to be on a streak, and it s attributed to his skill. 1) What is the probability that this player will get a hit exactly 5% of the time if he tries 0 times (i.e., 5 hits and 15 misses)? ) What is the probability that this player will get five hits (no misses) in a row? Lecture 6, p. 7

Solution The typical baseball player gets a hit 5% of the time. If this player gets several hits in a row, he is said to be on a streak, and it s attributed to his skill. 1) What is the probability that this player will get a hit exactly 5% of the time if he tries 0 times (i.e., 5 hits and 15 misses)? The probability of obtaining a particular microstate (five hits and 15 misses, in a specific order) is 0.5 5 0.75 15 = 1.3 10-5. Now, count microstates (different orderings of hits and misses): N = 0! / (5! 15!) = 15,504. Each microstate is equally likely, so the probability is 1.3 10-5 x 15,504 = 0.0. Here s one microstate: MMMMMHMHMMMHHMMMMHMM ) What is the probability that this player will get five hits (no misses) in a row? There is only one way to do this (one microstate). The probability is: P = 0.5 5 = 0.00098. That s fairly small (about one in a thousand) for a particular player, but not unlikely to happen by chance somewhere on a particular day, if one remembers that there are more than 1000 at bats every day in major league baseball. Lecture 6, p. 8